1 Express \(\frac { ( x - 7 ) ( x - 2 ) } { ( x + 2 ) ( x - 1 ) ^ { 2 } }\) in partial fractions.
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Question 1:
Answer Marks
Guidance
Answer Marks
Guidance
\(\frac{(x-7)(x-2)}{(x+2)(x-1)^2} \equiv \frac{A}{x+2} + \frac{B}{(x-1)} + \frac{C}{(x-1)^2}\) B1
If no partial fractions seen anywhere, B0. SC: \(\frac{A}{x+2} + \frac{Bx+C}{(x-1)^2}\) also acceptable
\((x-7)(x-2) \equiv A(x-1)^2 + B(x+2)(x-1) + C(x+2)\) M1
Allow careless minor error but not algebraic method error; or any equivalent identity with values of \(x\) substituted or coefficients compared
\(A=4, B=-3, C=2\) or \(\frac{4}{x+2} - \frac{3}{x-1} + \frac{2}{(x-1)^2}\) ISW A1,1,1
SC: \(A=4, B=-3, C=5\) or \(\frac{4}{x+2} + \frac{-3x+5}{(x-1)^2}\) gives max 3/5 for easier case
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# Question 1:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{(x-7)(x-2)}{(x+2)(x-1)^2} \equiv \frac{A}{x+2} + \frac{B}{(x-1)} + \frac{C}{(x-1)^2}$ | B1 | If no partial fractions seen anywhere, B0. SC: $\frac{A}{x+2} + \frac{Bx+C}{(x-1)^2}$ also acceptable |
| $(x-7)(x-2) \equiv A(x-1)^2 + B(x+2)(x-1) + C(x+2)$ | M1 | Allow careless minor error but not algebraic method error; or any equivalent identity with values of $x$ substituted or coefficients compared |
| $A=4, B=-3, C=2$ or $\frac{4}{x+2} - \frac{3}{x-1} + \frac{2}{(x-1)^2}$ ISW | A1,1,1 | SC: $A=4, B=-3, C=5$ or $\frac{4}{x+2} + \frac{-3x+5}{(x-1)^2}$ gives max 3/5 for easier case |
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1 Express $\frac { ( x - 7 ) ( x - 2 ) } { ( x + 2 ) ( x - 1 ) ^ { 2 } }$ in partial fractions.
\hfill \mbox{\textit{OCR C4 2013 Q1 [5]}}